K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a) Ta có: \(\sqrt{2021}-\sqrt{2020}\)

\(=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)

\(=\frac{1}{\sqrt{2020}+\sqrt{2021}}\)

Ta có: \(\sqrt{2020}-\sqrt{2019}\)

\(=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)

\(=\frac{1}{\sqrt{2019}+\sqrt{2020}}\)

Ta có: \(\sqrt{2020}+\sqrt{2021}>\sqrt{2019}+\sqrt{2020}\)

\(\Leftrightarrow\frac{1}{\sqrt{2020}+\sqrt{2021}}< \frac{1}{\sqrt{2019}+\sqrt{2020}}\)

hay \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)

b) Ta có: \(\sqrt{2019\cdot2021}\)

\(=\sqrt{\left(2020-1\right)\left(2020+1\right)}\)

\(=\sqrt{2020^2-1}\)

Ta có: \(2020=\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

nên \(\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow\sqrt{2019\cdot2021}< 2020\)

c) Ta có: \(\left(\sqrt{2019}+\sqrt{2021}\right)^2\)

\(=2019+2021+2\cdot\sqrt{2019\cdot2021}\)

\(=4040+2\sqrt{2019\cdot2021}\)

\(=4040+2\cdot\sqrt{2020^2-1}\)

Ta có: \(\left(2\sqrt{2020}\right)^2\)

\(=4\cdot2020\)

\(=4040+2\cdot2020\)

\(=4040+2\cdot\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)

\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)

Tham khảo:

loading...

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)

28 tháng 12 2020

Vì 2019 + 2020 < 2019 + 2021 nên A < B

24 tháng 5 2020

Ta có: \(\frac{2019}{2020}>\frac{2019}{2020+2021};\frac{2020}{2021}>\frac{2020}{2020+2021}\)

=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)

=> A > B.

9 tháng 8 2019

bài 1:

ssh của A là:

(151-3):2+1=75

A=(151+3)x75:2=5775

đáp số: 5775

14 tháng 5 2023

 

14 tháng 5 2023

oki

 

30 tháng 7 2020

Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)

=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)

Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)

=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)

Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)

=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)

=> 10B < 10A

=> B < A

b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)

Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)

=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> B < A

13 tháng 2 2022

sai rồi

5 tháng 1 2020

Xét 2017 /2018 và 2018/2019

1-2017/2018=1/2018

1-2018/2019=1/2019

mà 1/2018>1/2019=>2017/2018<2018/2019

Tương tự có:2020/2019>2021/2020

=>2017/2018+2010/2019<2018/2019+2021/2020

4 tháng 5 2021

Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A

 

4 tháng 5 2021

Thank you☺